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a(n) = Sum_{k=0..n} binomial(n+k,k)*sin(Pi*(n+k)/2).
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%I #31 Nov 11 2024 20:30:04

%S 0,1,-3,9,-30,106,-385,1421,-5304,19966,-75658,288222,-1102790,

%T 4234868,-16312773,63003869,-243896960,946066678,-3676303578,

%U 14308370014,-55768166380,217640082188,-850345208538,3325907590274,-13020993588680

%N a(n) = Sum_{k=0..n} binomial(n+k,k)*sin(Pi*(n+k)/2).

%H G. C. Greubel, <a href="/A181933/b181933.txt">Table of n, a(n) for n = 0..1000</a>

%F G.f.: (1/2)*(sqrt(4*x+1)*(1+x)-3*x-1)/(sqrt(4*x+1)*(x^2+3*x+1)-4*x^2-5*x-1). - _Vladimir Kruchinin_, Mar 28 2016

%F a(n) ~ (-1)^(n+1) *2^(2*n+1) / (5*sqrt(Pi*n)). - _Vaclav Kotesovec_, Mar 28 2016

%F Conjecture: +2*n*a(n) +8*n*a(n-1) +(-n+20)*a(n-2) +5*(-n+4)*a(n-3) +2*(-2*n+5)*a(n-4)=0. - _R. J. Mathar_, Jun 14 2016

%t f[n_] := Sum[ Binomial[n + k, k] Sin[Pi (n + k)/2], {k, 0, n}]; Array[f, 25, 0]

%o (Maxima)

%o makelist(coeff(taylor(1/2*(sqrt(4*x+1)*(1+x)-3*x-1)/(sqrt(4*x+1)*(x^2+3*x+1)-4*x^2-5*x-1),x,0,20),x,n),n,0,20); /* _Vladimir Kruchinin_, Mar 28 2016 */

%o (PARI) x='x+O('x^50); concat([0], Vec((1/2)*(sqrt(4*x+1)*(1+x)-3*x-1)/(sqrt(4*x+1)*(x^2+3*x+1)-4*x^2-5*x-1))) \\ _G. C. Greubel_, Mar 24 2017

%Y Cf. A026641, A176332.

%K sign

%O 0,3

%A _Robert G. Wilson v_, Apr 02 2012